Difference between revisions of "2021 Fall AMC 12A Problems/Problem 5"
MRENTHUSIASM (talk | contribs) (→Solution 2: Removed repetitive solution. Prof. Chen agreed to this through PM ...) |
MRENTHUSIASM (talk | contribs) (→Solution 1) |
||
| Line 6: | Line 6: | ||
<math>\textbf{(A) }6\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }15</math> | <math>\textbf{(A) }6\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }15</math> | ||
| − | == Solution | + | == Solution == |
There are <math>41-1=40</math> gaps between the <math>41</math> telephone poles, so the distance of each gap is <math>5280\div40=132</math> feet. | There are <math>41-1=40</math> gaps between the <math>41</math> telephone poles, so the distance of each gap is <math>5280\div40=132</math> feet. | ||
Revision as of 00:48, 26 November 2021
- The following problem is from both the 2021 Fall AMC 10A #6 and 2021 Fall AMC 12A #5, so both problems redirect to this page.
Problem
Elmer the emu takes 
equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in 
equal leaps. The telephone poles are evenly spaced, and the 
st pole along this road is exactly one mile (
feet) from the first pole. How much longer, in feet, is Oscar's leap than Elmer's stride?
Solution
There are 
gaps between the telephone poles, so the distance of each gap is 
feet.
Each of Oscar's leap covers 
feet, and each of Elmer's strides covers 
feet. Therefore, Oscar's leap is 
feet longer than Elmer's stride.
~MRENTHUSIASM
See Also
| 2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
| 2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
